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25 tháng 7 2019

Đề bài là j bạn

17 tháng 8 2016

1)x2-8x-9

= x^2 - 9x +x -9

= x(x+1) - 9 (x+1)

= (x-9) (x+1)

2)x2+3x-18

3)x3-5x2+4x

=x^3 - 4x^2 - x^2 + 4x 

= x^2 (x-1) - 4x(x-1)

= (x^2 - 4x) (x-1)

= x(x-4)(x-1)

4)x3-11x2+30x

5)x3-7x-6

6)x16-64

\(=\left(x^8\right)^2-8^2\)

\(=\left(x^8-8\right)\left(x^8+8\right)\)

7)x3-5x2+8x-4

8)x2-3x+2

= x^2 - 2x - x +2

= x(x-1) -2(x-1)

= (x-2)(x-1)

17 tháng 8 2016

1)   \(\left(x-9\right)\left(x+1\right)\)             2)   \(\left(x-3\right)\left(x+6\right)\)                                           3)   \(x\left(x-4\right)\left(x-1\right)\)

4)    \(x\left(x-6\right)\left(x-5\right)\)         5)\(\left(x-3\right)\left(x+1\right)\left(x+2\right)\)                               6)   ........

7)  \(\left(x-1\right)\left(x-2\right)\left(x-2\right)\)          8)   \(\left(x-2\right)\left(x-1\right)\)

2 tháng 4 2020

a. x3+5x2+3x-9

= x3-x2+6x2-6x+9x-9

= x2(x-1)+6x(x-1)+9(x-1)

= (x2+6x+9)(x-1)

= (x+3)2(x-1)

b. x3+9x2+11x-21

= x3-x2+10x2-10x+21x-21

= x2(x-1)+10x(x-1)+21(x-1)

= (x2+10x+21)(x-1)

= (x+7)(x+3)(x-1)

c. x3-7x+6

= x3-x2+x2-x-6x+6

= x2(x-1)+x(x-1)-6(x-1)

= (x2+x-6)(x-1)

= (x+3)(x-2)(x-1)

d. x3-5x2+8x-4

= x3-x2-4x2+4x+4x-4

= x2(x-1)-4x(x-1)+4(x-1)

= (x2-4x+4)(x-1)

= (x-2)2(x-1)

e. x3-3x+2

= x3+2x2-2x2-4x+x+2

= x2(x+2)-2x(x+2)+(x+2)

= (x2-2x+1)(x+2)

= (x-1)2(x+2)

f. x3+8x2+17x+10

= x3+5x2+3x2+15x+2x+10

= x2(x+5)+3x(x+5)+2(x+5)

= (x2+3x+2)(x+5)

= (x+1)(x+2)(x+5)

g. x3+3x2+6x+4

= x3+x2+2x2+2x+4x+4

= x2(x+1)+2x(x+1)+4(x+1)

= (x2+2x+4)(x+1)

h. x3-2x-4

= x3-2x2+2x2-4x+2x-4

= x2(x-2)+2x(x-2)+2(x-2)

= (x2+2x+2)(x-2)

k. x3+x2+4

= x3+2x2-x2-2x+2x+4

= x2(x+2)-x(x+2)+2(x+2)

= (x2-x+2)(x+2)

l. x3-12x+7x-2

= x3+2x2-2x2-4x-x-2

= x2(x+2)-2x(x+2)-(x+2)

= (x2-2x-1)(x+2)

2 tháng 4 2020

thansk you

1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)

\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)

\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)

\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)

\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)

Vậy: S={0;-7;8;-1}

2) Ta có: \(x^3-8x^2+17x-10=0\)

\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)

\(\Leftrightarrow x^2\left(x-2\right)-6x\left(x-2\right)+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-x-5x+5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=5\end{matrix}\right.\)

Vậy: S={2;1;5}

3) Ta có: \(2x^3-5x^2-x+6=0\)

\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)

\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+2x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-3\right)+\left(2x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)

Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)

4) Ta có: \(4x^4-4x^2-3=0\)

\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)

\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)

\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)

\(2x^2+1>0\forall x\in R\)

nên \(2x^2-3=0\)

\(\Leftrightarrow2x^2=3\)

\(\Leftrightarrow x^2=\frac{3}{2}\)

hay \(x=\pm\sqrt{\frac{3}{2}}\)

Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)

16 tháng 10 2016

\(2x^2+3x-27=2x^2-6x+9x-27=2x\left(x-3\right)+9\left(x-3\right)=\left(2x+9\right)\left(x-3\right)\)

\(x^3-7x+6=x^3-x-6x+6=x\left(x^2-1\right)-6\left(x-1\right)=x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)=\left(x-1\right)\left(x^2+x-6\right)\)

\(x^3+5x^2+8x+4=x^3+x^2+4x^2+8x+4=x^2\left(x+1\right)+4\left(x^2+2x+1\right)=x^2\left(x+1\right)+4\left(x+1\right)^2\)

\(=\left(x+1\right)\left(x^2+4x+4\right)=\left(x+1\right)\left(x+2\right)^2\)

\(27x^3-27x^2+18x-4=27x^3-9x^2-18x^2+6x+12x-4\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)

\(x^3+2x-3\)

\(=x^3-x+3x-3\)

\(=x\left(x^2-1\right)+3\left(x-1\right)\)

\(=x\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x\left(x+1\right)+3\right)\)

\(x^3+5x^2+8x+4\)

\(=\left(x+1\right)\left(x+2\right)\)

Cây cuối tương tự câu đầu thôi:

 \(x^3-7x+6\)

\(=x^3-x-6x+6\)

.......

9 tháng 8 2016

a)=-4x2+8x-4

=-[(2x)2-8x+4]

=-(2x-2)2

9 tháng 8 2016

b)=x3-3x2+3x2-9x+2x-6

=x2(x-3)+3x(x-3)+2(x-3)

=(x-3)(x2+3x+2)

=(x-3)(x2+x+2x+2)

=(x-3)[x(x+1)+2(x+1)]

=(x-3)(x+1)(x+2)

12 tháng 8 2018

1, x3+ 6x2+11x+6

= x3 + 2x2 + 4x2 + 8x + 3x + 6 

= x2(x + 2) + 4x(x + 2) + 3(x + 2)

= (x + 2)(x2 + 4x + 3)

2, x4+3x3-7x2-27x-18

= x4 + 3x3 - 9x2 + 2x2 - 27x -18

= (x4 - 9x2) + (3x3 - 27x) + (2x2 - 18)

= x2(x2 - 9) + 3x(x2 - 9) + 2(x2 - 9)

= (x2 - 9)(x2 + 3x + 2)

= (x + 3)(x - 3)(x2 + 3x + 2)

3, x3-8x2+x+42

= x3 - 7x2 - x2 + 7x - 6x + 42

= (x3 - 7x2) - (x2 - 7x) - (6x - 42)

= x2(x - 7) - x(x - 7) - 6(x - 7)

= (x - 7)(x2 - x - 6) 

4, x4+5x3-7x2-41x-30 

= x4 + x3 + 4x3 - 4x2 - 11x2 - 11x - 30x - 30

= (x4 + x3) + (4x3 - 4x2) - (11x2 + 11x) - (30x + 30)

= x3(x + 1) + 4x2(x + 1) - 11x(x + 1) - 30(x + 1)

= (x3 + 4x2 - 11x - 30)(x + 1)

5, x5+x-1

= x- x+ x+ x- x+ x- x2+ x -1 

= x3(x- x + 1)+ x2(x- x + 1)- (x- x + 1) 

= (x- x + 1)(x+ x- 1)

6, x5-x4-1

= x5 - x3 - x2 - x4 + x2 + x + x3 - x - 1 

= x2(x3 - x - 1) - x(x3 - x - 1) + (x3 - x - 1)

= (x2 - x + 1)(x3 - x - 1)

12 tháng 8 2018

1, x 3+ 6x 2+11x+6

= x 3 + 2x 2 + 4x 2 + 8x + 3x + 6

= x 2 ﴾x + 2﴿ + 4x﴾x + 2﴿ + 3﴾x + 2﴿

= ﴾x + 2﴿﴾x 2 + 4x + 3﴿

2, x 4+3x 3‐7x 2‐27x‐18

= x 4 + 3x 3 ‐ 9x 2 + 2x 2 ‐ 27x ‐18

= ﴾x 4 ‐ 9x 2 ﴿ + ﴾3x 3 ‐ 27x﴿ + ﴾2x 2 ‐ 18﴿

= x 2 ﴾x 2 ‐ 9﴿ + 3x﴾x 2 ‐ 9﴿ + 2﴾x 2 ‐ 9﴿

= ﴾x 2 ‐ 9﴿﴾x 2 + 3x + 2﴿

=﴾x + 3﴿﴾x ‐ 3﴿﴾x 2 + 3x + 2﴿

3, x 3‐8x 2+x+42

= x 3 ‐ 7x 2 ‐ x 2 + 7x ‐ 6x + 42

= ﴾x 3 ‐ 7x 2 ﴿ ‐ ﴾x 2 ‐ 7x﴿ ‐ ﴾6x ‐ 42﴿

= x 2 ﴾x ‐ 7﴿ ‐ x﴾x ‐ 7﴿ ‐ 6﴾x ‐ 7﴿

= ﴾x ‐ 7﴿﴾x 2 ‐ x ‐ 6﴿

4, x 4+5x 3‐7x 2‐41x‐30

= x 4 + x 3 + 4x 3 ‐ 4x 2 ‐ 11x 2 ‐ 11x ‐ 30x ‐ 30

= ﴾x 4 + x 3 ﴿ + ﴾4x 3 ‐ 4x 2 ﴿ ‐ ﴾11x 2 + 11x﴿ ‐ ﴾30x + 30﴿

= x 3 ﴾x + 1﴿ + 4x 2 ﴾x + 1﴿ ‐ 11x﴾x + 1﴿ ‐ 30﴾x + 1﴿

= ﴾x 3 + 4x 2 ‐ 11x ‐ 30﴿﴾x + 1﴿

5, x 5+x‐1

= x 5 ‐ x 4 + x 3 + x 4 ‐ x 3 + x 2 ‐ x 2+ x ‐1

= x 3 ﴾x 2 ‐ x + 1﴿+ x 2 ﴾x 2 ‐ x + 1﴿‐ ﴾x 2 ‐ x + 1﴿

= ﴾x 2 ‐ x + 1﴿﴾x 3 + x 2 ‐ 1﴿ 6, x 5‐x 4‐1

= x 5 ‐ x 3 ‐ x 2 ‐ x 4 + x 2 + x + x 3 ‐ x ‐ 1

= x 2 ﴾x 3 ‐ x ‐ 1﴿ ‐ x﴾x 3 ‐ x ‐ 1﴿ + ﴾x 3 ‐ x ‐ 1﴿

= ﴾x 2 ‐ x + 1﴿﴾x 3 ‐ x ‐ 1﴿ 

28 tháng 1 2020

a. x3+4x2-29x+24